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IB Review

IB Review. Following the Data Packet The front matter Section by section formulas/problems Linear kinematics Dynamics (F = ma) Circular Motion Energy Momentum Waves Thermal Field Theory Currents and Induction The rest of the formulas. Uncertainty and vector components.

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IB Review

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  1. IB Review Following the Data Packet • The front matter • Section by section formulas/problems • Linear kinematics • Dynamics (F = ma) • Circular Motion • Energy • Momentum • Waves • Thermal • Field Theory • Currents and Induction • The rest of the formulas

  2. Uncertainty and vector components

  3. Linear Kinematics v = u + at ?????

  4. An air rocket leaves the ground straight up, and strikes the ground 4.80 seconds later. • What time does it take to get to the top? • How high does it go? • What was its initial velocity? • What is the velocity at elevation 21.0 m? 2.4 s, 28.2 m, 23.5 m/s, + or - 11.9 m/s

  5. H V s u v a t s u v a t Av = A sin  AH = A cos 2-Dimensional Motion Pythagorean x2 + y2 = hyp2

  6. V = 9.21 m/s t = 2.17 s • How far out does she land? • How high is the cliff? • What is the velocity of impact in VC Notation? • What is the velocity of impact? (in AM Notation) • 20.0 m, 23.1 m, • 9.21 m/s x + -21.3 m/s y • 23.2 m/s 66.6o below horiz

  7. Find vector components • Fill in your H/V table of suvat • Find the horizontal distance traveled • Find velocity of impact in angle magnitude v = 126 m/s angle = 43.0o The cliff is 78.5 m tall 1690 m, 133 m/s@ 46.3o

  8. Dynamics

  9. Find the force: F = ma, wt = 1176 N downward <F – 1176 N> = (120. kg)(-4.50 m/s/s) F – 1176 N = -540 N F = 636 N … F 120. kg a = -4.50 m/s/s (DOWNWARD) 636 N

  10. A 120 mW laser uses a wavelength of 656 nm. What is the energy and momentum of a photon of light at this wavelength? How many photons per second does it emit? What force would it exert on an object that absorbs the photons? How would that change if the photons were reflected? 3.030E-19 J, 1.010E-27 kg m/s, 3.960E17 photons/sec, 4.00E-10 N

  11. Also on page 8: Gravity and Circular Motion

  12. A Volkswagen can do .650 “g”s of lateral acceleration. What is the minimum radius turn at 27.0 m/s? (3) a = v2/r 1g= 9.81 m/s/s a = (9.81 m/s/s)(.650) = 6.3765 m/s/s 6.3765 m/s/s = (27.0 m/s)2/r r = (27.0 m/s)2/(6.3765 m/s/s) = 114.326 m r = 114m 114m

  13. What should be the period of motion if you want 3.5 “g”s of centripetal acceleration 5.25 m from the center of rotation? a = 42r/T2 a = (3.5)(9.8 m/s/s) = 34.3 m/s/s 34.3 m/s/s = 42(5.25 m)/T2 T = 2.5 s … 2.5 s

  14. Ice skates can give 420 N of turning force. What is rmin for a 50.kg skater @10.m/s? F=ma, a=v2/r F=mv2/r 420 N = (50 kg)(10.m/s)2/r r = (50 kg)(10.m/s)2/(420 N) r = 11.9m 11.9m

  15. The moon has a mass of 7.36 x 1022 kg, and a radius of 1.74 x 106 m. What does a 34.2 kg mass weight on the surface? • r = Center to center distance • m1 = One of the masses • m2 = The other mass • G = 6.67 x 10-11 Nm2/kg2 • F = Gm1m2 • r2 • F = 55.5 N • … 55.5 N

  16. msv2 r • = Gmsmc • r2 At what distance from the moon’s center is the orbital velocity 52.5 m/s? Mm = 7.36 x 1022 kg • r = Gmc • v2 1781086621 m 1.78 x 109 m

  17. Also Power = work/time Eelas = 1/2kx2 Energy

  18. What speed at the bottom? vi = 5.8 m/s 15 kg h = 2.15 m Fd + mgh + 1/2mv2 =Fd + mgh + 1/2mv2 0 + mgh + 1/2mv2 =0 + 0 + 1/2mv2 (15 kg)(9.8 N/kg)(2.15 m) + 1/2(15 kg)(5.8 m/s)2= 1/2(15 kg)v2 v = 8.7 m/s … 8.7 m/s

  19. Ima Wonder can put out 127 W of power. What time will it take her to do 671 J of work? P = W/t, t = W/P = (671 J)/(127 W) = 5.28 s 5.28 s

  20. Frieda People can put out 430. W of power. With what speed can she push a car if it takes 152 N to make it move at a constant velocity? P = Fv v = P/F = (430. W)/(152 N) = 2.83 m/s 2.83 m/s

  21. What must be the power rating of a motor if it is to lift a 560 kg elevator up 3.2 m in 1.5 seconds? 11700 W

  22. Momentum

  23. Jolene exerts a 50. N force for 3.00 seconds on a stage set. It speeds up from rest to .25 m/s. What is the mass of the set? (m)(v) = (F )( t) (m)(.25 m/s) = (50. N )(3.0 s) m = (50. N )(3.0 s)/(.25 m/s) = 600 kg = 6.0 x102 kg … 6.0 x102 kg

  24. + Before After 6.20m/s 1.20 m/s v = ? 17.0 kg 17.0 kg 13.0 kg 13.0 kg (13kg+17kg)(6.2m/s) = (13kg)(-1.2m/s)+(17kg)v 186kgm/s = -15.6kgm/s+(17kg)v 201.6kgm/s = (17kg)v (201.6kgm/s)/(17kg) = 11.9 m/s = v … 11.9 m/s

  25. Topic 4: Oscillations and waves

  26. Oscillations and waves

  27. Simple Harmonic Motion - Kinematics  = 2f = 1 = 2f T T x = xosin(t) or xocos(t) v = vocos(t) or -vosin(t)  – “Angular” velocity T – Period of motion x – Position (at some time) v – Velocity (at some time) Draw on board: xo – Max Position (Amplitude) vo – Max Velocity

  28. xo = Maximum displacement • (AKA Amplitude) • vo = Maximum velocity • ao = Maximum acceleration • x: -xo 0 +xo • v: 0 +/-vo 0 • a: +ao 0 -ao

  29. Simple Harmonic Motion - Energy Ek = 1/2m2(xo2 – x2) Ek (max) = 1/2m2xo2 ET = 1/2m2xo2 ET – Total Energy Ek (max) – Maximum Kinetic Energy Ek – Kinetic Energy  – “Angular” velocity T – Period of motion x – Position (at some time) v – Velocity (at some time) xo – Max Position (Amplitude) vo – Max Velocity

  30. Derive the energy equations: Ek = 1/2m2(xo2 – x2) Ek (max) = 1/2m2xo2 ET = 1/2m2xo2 Simple Harmonic Motion - Energy Ek (max) = 1/2mvo2 Ep (max) = 1/2kxo2 Where they happen • Ek: 0 max 0 • Ep: max 0 max

  31. What is the period of a guitar string that is vibrating 156 times a second? (156 Hz) Use f = 1/T W 0.00641 s

  32. A mass on the end of a spring oscillates with a period of 2.52 seconds and an amplitude of 0.450 m. What is its maximum velocity? (save this value) v = + ( xo2- x2), make x = 0,  = 2/2.52, |v| = 1.12199…. m/s W 1.12 m/s

  33. A SHO has an equation of motion of: (in m) x = 2.4sin(6.1t) a) what is the amplitude and angular velocity of the oscillator? b) what is its period? c) what is its maximum velocity? d) write an equation for its velocity. xo = 2.4 m,  = 6.1 rad/s T = 2/6.1 = 1.03 s vo = (6.1 rad/s)(2.4 m) = 14.64 v = 15cos(6.1t) 2.4 m – 6.1 rad/s 1.0 s 15 m/s v = 15cos(6.1t) W

  34. A loudspeaker makes a pure tone at 440.0 Hz. If it moves with an amplitude of 0.87 cm, what is its maximum velocity? (0.87 cm = .0087 m) (f = 1/T) v = + ( xo2- x2), make x = 0,  = 2(440), |v| = 24.052…. m/s W 24 m/s

  35. A mass on the end of a spring oscillates with a period of 1.12 seconds and an amplitude of 0.15 m. Suppose it is moving upward and is at equilibrium at t = 0. What is its velocity at t = 13.5 s? use v = vocos(t),  = 2/1.12, vo =  ( xo2) = xo, v = +0.79427… m/s W +0.79 m/s

  36. An SHO has a mass of 0.259 kg, an amplitude of 0.128 m and an angular velocity of 14.7 rad/sec. What is its total energy? (save this value in your calculator) Use ET = 1/2m2xo2 W 0.458 J

  37. An SHO has a mass of 0.259 kg, an amplitude of 0.128 m and an angular velocity of 14.7 rad/sec. What is its kinetic energy when it is 0.096 m from equilibrium? What is its potential energy? Use Ek = 1/2m2(xo2 – x2) W 0.20 J, 0.26 J

  38. An SHO has a total energy of 2.18 J, a mass of 0.126 kg, and a period of 0.175 s. a) What is its maximum velocity? b) What is its amplitude of motion? Use Ek = 1/2mv2 Then  = 2/T Use Ek (max) = 1/2m2xo2 W 5.88 m/s J, 0.164 m

  39. An SHO a maximum velocity of 3.47 m/s, and a mass of 0.395 kg, and an amplitude of 0.805 m. What is its potential energy when it is 0.215 m from equilibrium?  = 2/T Use Ek = 1/2mv2 Use Ek (max) = 1/2m2xo2 Then Use Ek = 1/2m2(xo2 – x2) Subtract kinetic from max W 0.170 J

  40. A 1250 kg car moves with the following equation of motion: (in m) • x = 0.170sin(4.42t) • a) what is its total energy? • b) what is its kinetic energy at t = 3.50 s? Use ET = 1/2m2xo2 Then find x from the equation: (.04007…) Then use Use Ek = 1/2m2(xo2 – x2) W 353 J, 333 J

  41. Oscillations and waves

  42. What is the frequency of a sound wave that has a wavelength of 45 cm, where the speed of sound is 335 m/s v = f f = v/ = (335 m/s)/(.45 m) = 744.444 = 740 Hz … 740 Hz

  43. The waveform is 62 cm long. What is the ? If it is a sound wave (v = 343 m/s), what is its frequency (v = f) L = 2/4  = 4/2(.62 m) = 1.24 m v = f, f = v/ = (343 m/s)/(1.24 m) = 277 Hz … 277 Hz

  44. The waveform is 2.42 m long. What is the ? If it is a sound wave (v = 343 m/s), what is its frequency (v = f) L = 1/4  = 4/1(2.42 m) = 9.68 m v = f, f = v/ = (343 m/s)/(9.68 m) = 35.4 Hz ... 35.4 Hz

  45. Oscillations and waves

  46. A person who is late for a concert runs at 18.0 m/s towards an A 440.0 Hz. What frequency do they hear? (use v sound = 343 m/s) • Moving observer • higher frequency • f’ = f{1 + vo/v} • f = 440.0 Hz, vo = 18.0 m/s, v = 343 m/s, and + • F = 463 Hz • … 463 Hz

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